bạn bt rùi sao còn kêu tụi mình cm chi nữa

mk đã bk dou bn

\(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2.n^2+2n+1}< \frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{2.n^2+2n}\)

\(A< \frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n.\left(n+1\right)}\right)\)

\(A< \frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}\right)\)

\(A< \frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{n}-\frac{1}{n+1}\right)\)

\(A< \frac{1}{2}.\left(1-\frac{1}{n+1}\right)< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2}\)

Ta có: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\)

\(\Rightarrow\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}\)

\(=\frac{1}{1^2+2^2}+\frac{1}{2^2+3^2}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2.1.2}+\frac{1}{2.2.3}+...+\frac{1}{2.n.\left(n+1\right)}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{n.\left(n+1\right)}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{n+1}\right)< \frac{1}{2}\)

\(VP=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot5}+\dfrac{1}{5\cdot13}+\dfrac{1}{13\cdot8}+\dfrac{1}{8\cdot19}+\dfrac{1}{19\cdot11}+\dfrac{1}{11\cdot25}\\ =\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{10\cdot13}+\dfrac{2}{13\cdot16}+\dfrac{2}{16\cdot19}+\dfrac{2}{19\cdot22}+\dfrac{2}{22\cdot25}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}+\dfrac{3}{19\cdot22}+\dfrac{3}{22\cdot25}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\left(1-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\dfrac{24}{25}\\ =\dfrac{16}{25}\)\(a^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\\ a^2-\dfrac{9}{25}=\dfrac{16}{25}\\ a^2=\dfrac{16}{25}+\dfrac{9}{25}\\ a^2=1\\ \Rightarrow\left[{}\begin{matrix}a=1\\a=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-a=-1\\-a=1\end{matrix}\right.\)

Vậy \(-a=-1\) hoặc \(-a=1\)

\(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2.n^2+2n+1}< \frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{2.n^2+2n}\)

\(A< \frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n.\left(n+1\right)}\right)\)

\(A< \frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}\right)\)

\(A< \frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)

\(A< \frac{1}{2}.\left(1-\frac{1}{n+1}\right)< \frac{1}{2}\)

=> \(A< \frac{1}{2}\)

\(a^2+\left(a+1\right)^2=a^2+a^2+2a+1\\ =2a^2+2a+1>2a\left(a+1\right)\\ \Rightarrow\dfrac{1}{a^2+\left(a+1\right)^2}< \dfrac{1}{2a\left(a+1\right)}\)

\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^{^2}}\\ =\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\\ < \dfrac{1}{2.1.\left(1+2\right)}+\dfrac{1}{2.2\left(2+1\right)}+....+\dfrac{1}{2n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{n+1}\right)\\ =\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{n+1}\right)\\ =\dfrac{5}{12}-\dfrac{1}{2n+2}< \dfrac{5}{12}< \dfrac{9}{20}\)